IMO 2012 Shortlist G2

Category: Geometry

Problem

Let ABCD be a cyclic quadrilateral whose diagonals AC and BD meet at E. The extensions of the sides AD and BC beyond A and B meet at F. Let G be the point such that ECGD is a parallelogram, and let H be the image of E under reflection in AD. Prove that D, H, F, G are concyclic. Solution. We show first that the triangles FDG and FBE are similar. Since ABCD is cyclic, the triangles EAB and EDC are similar, as well as FAB and FCD. The parallelogram ECGD yields GD = EC and ∠CDG = ∠DCE; also ∠DCE = ∠DCA = ∠DBA by inscribed angles. Therefore ∠FDG = ∠FDC + ∠CDG = ∠FBA + ∠ABD = ∠FBE, GD EB

CE EB

CD AB

FD FB . It follows that FDG and FBE are similar, and so ∠FGD = ∠FEB. A B D G E F H C Since H is the reflection of E with respect to FD, we conclude that ∠FHD = ∠FED = 180◦ − ∠FEB = 180◦ − ∠FGD. This proves that D, H, F, G are concyclic. Comment. Points E and G are always in the half-plane determined by the line FD that contains B and C, but H is always in the other half-plane. In particular, DHFG is cyclic if and only if ∠FHD + ∠FGD = 180◦.31